document.write( "Question 888276: If f(x) = √x^2-1 and g(x) = √x-1 which expression represents f(x)/g(x) for x>1? \n" ); document.write( "
Algebra.Com's Answer #537265 by Theo(13342)\"\" \"About 
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\"sqrt%28x%5E2+-+1%29+%2F+sqrt%28x-1%29\" = \"sqrt%28%28x%5E2-1%29%2F%28x-1%29%29\" = \"sqrt%28%28x-1%29%28x%2B1%29%2F%28x-1%29%29\" = \"sqrt%28x%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "following is the graph of the equation of \"sqrt%28x%5E2+-+1%29+%2F+sqrt%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "following is the graph of the equation of \"sqrt%28x%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "the first graph doesn't allow values of x = 1 because that would make the denominator equal to 0 and the function would be undefined.\r
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\n" ); document.write( "\n" ); document.write( "the first graph also doesn't allow values of x <1 because that would make the numerator the squart root of a negative number which is also undefined in the real number system.\r
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\n" ); document.write( "\n" ); document.write( "the second graph goes to 0 because x+1 will be positive for values less than 1 of x but greater than or equal to 0 and will also allows x = 1 because there is no denominator in that equation.\r
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\n" ); document.write( "\n" ); document.write( "the 2 equations are equivalent except that once you eliminated the denominator by canceling it out, you can then get a defined solution at x = 1.\r
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